pointer to an array int (*ptr)[]

Question

I am trying to understand how a pointer to an array works. A code snippet;

#include<stdio.h> 
  
int main() 
{ 
  int arr[3] = { 0 , 8 ,10 };
  int (*ptr)[3] = &arr;
  int i = 0;
  for (i = 0; i < 3 ; i++)
  printf("Address (%p) - value( %d)\n", (*ptr+i) , *(*ptr + i));
  return 0; 
} 

An asterisk * derefernces the ptr . If i = 1, why is (*ptr+i) = ith value not value at ptr + i.

Answer 1

The type of ptr is int (*)[3] (pointer to an array length 3 of int). The type of *ptr is int[3] (array length 3 of int). In most expressions, an operand of type int[3] is converted to a int * (pointer to int) pointing to the first element of the array. The expression (*ptr+i) results in a pointer to the ith element of the array by pointer arithmetic. In the expression *(*ptr+i), the pointer to the ith element of the array is dereferenced to produce the value of the ith element of the array, which is of type int.

Answer 2

Here's how the expressions relate to each other:

   ptr  ==  &arr  // int (*)[3] == int (*)[3]
  *ptr  ==   arr  // int [3]    == int [3]

Unless it is the operand of the sizeof or unary & operators, or is a string literal used to initialize a character array in a declaration, an expression of type "N-element array of T" will be converted, or "decay", to an expression of type "pointer to T" and the value of the expression will be the address of the first element of the array. Thus in the expressions below, *ptr and arr "decay" from type "3-element array of int" to "pointer to int", and their values are the address of the first element of arr:

 (*ptr + i)  ==  (arr + i) == &arr[i]  // int * == int * == int *
*(*ptr + i)  == *(arr + i) ==  arr[i]  // int   == int   == int 

Answer 3

This one uses malloc, but the concept is same as assigning the address of 1-D array.

read the comments inline , for better understanding.

#include <stdio.h>
#include <stdlib.h>

#define NCOLS 5
#define NROWS 2

void function( int (*_2D_ptr_Array)[NCOLS] , int nrows, int ncols );

//enable this to see the demo of accessing elements using pointer to an array 
//disabling this will demo how to pass 2-D array and collect using pointer to an array.

//#define _DEMO_MAIN 

int main()
{
    // a is a pointer to an array of 5 ints
    int (*a)[NCOLS];
    int _2D_Array[NROWS][NCOLS] =   {   
                                        { 10, 20, 30, 40, 50},
                                        { 60, 70, 80, 90, 100}
                                    };
    int i,j;
    
    #ifdef _DEMO_MAIN
    printf("Enter 10 elements\n");
    
    a = malloc(sizeof(int)*5*2);
    
    printf("a got address at %p\n", a);
    
    for(i=0;i<2;i++)
        for(j=0;j<5;j++)
            scanf("%d",a[i]+j);

    //printing one dimentional array values
    // *a is equivalent to a[j], when j = 0
    for(j=0;j<5;j++)
        printf("value  %d is at addr %p\n",*(*a+j),(*a+j));
    
    putchar('\n');
    
    //printing multi-dimentional array values
    // adding j to a[i] or *a gives the next element address in a given row.
    for(i=0;i<2;i++)
        for(j=0;j<5;j++)
            printf("value %d is at %p\n",*(a[i]+j), a[i]+j);
    
    #else
        //a most common use of pointer to an array is to collect the 2-d array as an argument in function
        function( _2D_Array, NROWS, NCOLS );
    #endif
    
    return 0;
}

void function( int (*_2D_ptr_Array)[NCOLS], int nrows, int ncols ) {
    for(int r = 0; r < nrows ; r++) {
        for(int c = 0; c < ncols ; c++) 
            printf("%d ", _2D_ptr_Array[r][c]);
        printf("\n");
    }
}

Note: The return checks for the functions like scanf and malloc are not handled purposefully